Problem: You have found the following ages (in years) of all 6 zebras at your local zoo: $ 3,\enspace 13,\enspace 19,\enspace 27,\enspace 4,\enspace 13$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{3 + 13 + 19 + 27 + 4 + 13}{{6}} = {13.2\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $3$ years $-10.2$ years $104.04$ years $^2$ $13$ years $-0.2$ years $0.04$ years $^2$ $19$ years $5.8$ years $33.64$ years $^2$ $27$ years $13.8$ years $190.44$ years $^2$ $4$ years $-9.2$ years $84.64$ years $^2$ $13$ years $-0.2$ years $0.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{104.04} + {0.04} + {33.64} + {190.44} + {84.64} + {0.04}} {{6}} $ $ {\sigma^2} = \dfrac{{412.84}}{{6}} = {68.81\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{68.81\text{ years}^2}} = {8.3\text{ years}} $ The average zebra at the zoo is 13.2 years old. There is a standard deviation of 8.3 years.